  Question

# Let A=Q×Q, where Q is the set of all rational numbers, and * be a binary operation on A defined by (a, b) * (c, d) = (ac, b + ad) for (a.b),(c,d)ϵA. Then find (i) The identity element of * in A. (ii) Invertible elements of A, and hence write the inverse of elements (5, 3) and. (12,4). OR Let f : W→W be defined as f(n){n−1,if n is oddn+1,if n is even Show that f is invertible and find the inverse of f. Here, W is the set of all whole numbers.

Solution

## Let A=Q×Q, where Q is the set of rational numbers. Given that * is the binary operation on A defined by (a, b) * (c, d) = (ac, b + ad) for (a,b),(c,d)^iA. (i) We need to find the identity element of the operation * in A. Let (x, y) be the identity element in A. Thus, (a, b) * (x, y) = (x, y) * (a, b) = (a, b), for all (a,b)ϵA ⇒(ax,b+ay)=(a,b) ⇒ax=a and b+ay=b ⇒y=0 and x=1 Therefore, (1,0)ϵA is the identity element in A with respect to the operation *. (ii) We need to find the invertible elements of A. Let (p, q) be the inverse of the element (a, b) Thus, (a,b)∗(p,q)=(1,0) ⇒(ap,b+aq)=(1,0) ⇒ap=1 and b+aq=0 ⇒p=1aand q=−ba Thus the inverse elements of (a, b) is (1a,−ba) Now let us find the inverse of (5, 3) and (12,4) Hence, inverse of (5, 3) is (12,−35) And inverse of  (12,4)is(2,−412)=(2,−8) OR Let f: W→W be defined as f(n){n−1,if n is oddn+1,if n is even We need to prove that 'f' is invertible. In order to prove that 'f' is invertible it is sufficient to prove that f is a bijection. A function f: A→B is a one-one function or an injection, if f(x)=f(y)⇒x=y for all x,yϵA. Case i: If x and y are odd. Let f(x) = f(y) ⇒x−1=y−1 ⇒x=y Case ii: If x and y are even, Let f(x) = f(y) ⇒x+1=y+1 ⇒x=y Thus, in both the cases, we have, f(x)=f(y)→x=y for all x,yϵW. Hence f is an injection. Let n be an arbitrary element of W. If n is an odd whole number, there exists an even whole number n−1ϵW such that f(n - 1) = n − 1 + 1 = n. If n is an even whole number, then there exists an odd whole number n+1ϵW such that f(n + 1) = n + 1 − 1 = n. Also, f(1) = 0 and f(0) = 1 Thus, every element of W (co-domain) has its pre-image in W (domain). So f is an onto function. Thus, it is proved that f is an invertible function. Thus, a function g:B→A which associates each element yϵB to a unique element xϵA such that f(x) = y is called the inverse of f. That is, f(x)=y⇔g(y)=x The inverse of f is generally denoted by f−1. Now let us find the inverse of f. Let x,yϵW such that f(x) = y ⇒x+1=y, if x is even And x − 1 = y, if x is odd ⇒x={y−1,if y is oddy+1,if y is even ⇒f−1(y)={y−1,if y is oddy+1,if y is even Interchange, x and y, we have, ⇒f−1(x)={x−1,if x is oddx+1,if x is even144 Rewriting the above we have, ⇒f−1(x)={x−1,if x is evenx+1,if x is oddThus,f−1(x)=f(x)  Suggest corrections   